Equation of a Tangent Line to a Curve

Relationship Between Derivatives and Tangent Lines

You probably already know that derivatives can tell us a lot about the properties of a function. One of the coolest uses of derivatives is to find the slope or gradient of a tangent line to a curve.

Imagine "zooming in" on a point on a curve over and over. Eventually, the curved line will start to look like a straight line, right? Well, that imaginary straight line is what we call the tangent line. The gradient of the tangent line at a point $x=a$ on the curve $y=f(x)$ is exactly equal to the derivative of the function at that point, which is $m = f'(a)$.

Determining the Equation of a Tangent Line

To create the equation of a straight line, we need two main things: a point the line passes through and the gradient of the line itself. In this context:

1. Point of Tangency: A point $P(a, b)$ where the line touches the curve.
2. Gradient (m): The slope of the line at that point, which we get from the derivative, $m = f'(a)$.

Once we have both, we can plug them directly into the basic formula for a line equation that you're already familiar with:

In short, to find the equation of a tangent line, first find its gradient by differentiating the function, then plug the point of tangency and the gradient into the line equation formula.

Breaking Down a Case

Let's see how it works with an example. Determine the equation of the tangent line to the parabola $y = 2 - 4x^2$ at the point $(1, -2)$.

Solution:

Step 1: Find the gradient of the tangent line

First, we differentiate the function $f(x) = 2 - 4x^2$ to get its gradient expression.

Since we want to find the gradient at the point where the x-coordinate is $x=1$, we substitute this value into the derivative.

So, the gradient of the tangent line is -8.

Step 2: Construct the equation

Now we have everything we need:

• Point of tangency $(a, b) = (1, -2)$
• Gradient $m = -8$

Plug them into the line equation formula:

So, the equation of the tangent line is $y = -8x + 6$.

Exercises

1. Find the equation of the tangent line to the curve $y = x^2 - 6\frac{1}{5}x + 14\frac{1}{2}$ which is parallel to the line $x + 2y + 3 = 0$.

2. Determine the equation of the tangent line to the parabola $y = 2x^2 - 3x + 5$ at the point with an ordinate of 4.

3. A curve $y = 3x - \frac{3}{x^2}$ intersects the X-axis at P. Find the equation of the tangent line to the curve at point P!

4. The curve $y = (x^2 + 2)^2$ intersects the Y-axis at point A. Show that the tangent line to the curve at point A is parallel to the X-axis and is 4 units away from the origin!

5. Determine the coordinates of the point on the curve $y = 2x^2 - 7x + 1$, if the tangent line to the curve at that point forms an angle of 45° with the positive X-axis. Also, determine the equation of the tangent line to the curve that passes through that point!

Answer Key

1. Solution:

   Step 1: Determine the gradient.

   The tangent line must be parallel to the line $x + 2y + 3 = 0$. To find its gradient, let's first convert this line's equation into the slope-intercept form $y = mx + c$.

   $$2y = -x - 3$$

   $$y = -\frac{1}{2}x - \frac{3}{2}$$

   From this, we know the gradient of the line is $m = -\frac{1}{2}$. Since the tangent line is parallel, its gradient is the same.

   Step 2: Find the point of tangency.

   The gradient of the curve at a point is equal to the value of the first derivative at that point. First, let's find the derivative function of $f(x) = x^2 - \frac{31}{5}x + \frac{29}{2}$.

   $$f'(x) = 2x - \frac{31}{5}$$

   Next, we set this derivative equal to the gradient we know ($m = -1/2$) to find the x-coordinate of the point of tangency.

   $$2x - \frac{31}{5} = -\frac{1}{2}$$

   $$2x = \frac{31}{5} - \frac{1}{2} = \frac{62 - 5}{10} = \frac{57}{10}$$

   $$x = \frac{57}{20}$$

   After getting the x-coordinate $x=\frac{57}{20}$, we substitute this value back into the original curve equation to find its y-coordinate.

   $$y = (\frac{57}{20})^2 - \frac{31}{5}(\frac{57}{20}) + \frac{29}{2} = \frac{3249}{400} - \frac{1767}{100} + \frac{29}{2} = \frac{3249 - 7068 + 5800}{400} = \frac{1981}{400}$$

   So, the point of tangency is $\left(\frac{57}{20}, \frac{1981}{400}\right)$.

   Step 3: Construct the line equation.

   With the point $\left(\frac{57}{20}, \frac{1981}{400}\right)$ and gradient $m = -\frac{1}{2}$, the equation is:

   $$y - \frac{1981}{400} = -\frac{1}{2}\left(x - \frac{57}{20}\right)$$

   $$y = -\frac{1}{2}x + \frac{57}{40} + \frac{1981}{400}$$

   $$y = -\frac{1}{2}x + \frac{570 + 1981}{400}$$

   $$y = -\frac{1}{2}x + \frac{2551}{400}$$

2. Solution:

   Step 1: Find the point(s) of tangency.

   Since the ordinate is 4, we set $y=4$ in the parabola's equation.

   $$4 = 2x^2 - 3x + 5$$

   $$2x^2 - 3x + 1 = 0$$

   $$(2x-1)(x-1) = 0$$

   From factorization, we get two x-values: $x=1$ and $x=\frac{1}{2}$. This means there are two points of tangency: $(1, 4)$ and $\left(\frac{1}{2}, 4\right)$. Therefore, there will be two tangent line equations.

   Step 2: Calculate the gradient and create the equation for each point.

   We will process each point of tangency separately. The derivative of the function is $f'(x) = 4x - 3$.

   Case One: Point $(1, 4)$

   The gradient at this point is $m = f'(1) = 4(1) - 3 = 1$.

   Thus, the equation is:

   $$y - 4 = 1(x - 1) \implies y = x + 3$$

   Case Two: Point $\left(\frac{1}{2}, 4\right)$

   The gradient at this point is $m = f'(\frac{1}{2}) = 4(\frac{1}{2}) - 3 = -1$.

   Thus, the equation is:

   $$y - 4 = -1(x - \frac{1}{2}) \implies y = -x + \frac{9}{2}$$

3. Solution:

   Step 1: Find point P.

   The curve intersects the X-axis when $y=0$.

   $$0 = 3x - \frac{3}{x^2}$$

   $$3x = \frac{3}{x^2} \implies 3x^3 = 3 \implies x^3 = 1 \implies x=1$$

   So, the intersection point P is $(1, 0)$.

   Step 2: Find the gradient at P.

   Let's first rewrite the function as $f(x) = 3x - 3x^{-2}$ to make it easier to differentiate.

   $$f'(x) = 3 - (-2)(3)x^{-3} = 3 + \frac{6}{x^3}$$

   The gradient at $x=1$ is $m = f'(1) = 3 + \frac{6}{1^3} = 9$.

   Step 3: Construct the equation.

   With point $(1, 0)$ and gradient 9, the equation is:

   $$y - 0 = 9(x - 1) \implies y = 9x - 9$$

4. Solution:

   Step 1: Find point A.

   The curve intersects the Y-axis when $x=0$.

   $$y = (0^2 + 2)^2 = 4$$

   So, the intersection point A is $(0, 4)$.

   Step 2: Prove the tangent line is parallel to the X-axis.

   A line parallel to the X-axis must have a gradient of 0. Let's prove that the derivative of the function at point A ($x=0$) is zero.

   The derivative of $f(x)=(x^2+2)^2$ using the chain rule is $f'(x) = 2(x^2+2)(2x) = 4x(x^2+2)$.

   The gradient at $x=0$ is $m = f'(0) = 4(0)(0^2+2) = 0$.

   Since the gradient is zero, it is proven that the tangent line is parallel to the X-axis.

   Step 3: Prove its distance is 4 units from the origin.

   The equation of the tangent line at point $(0,4)$ with gradient $m=0$ is:

   $$y - 4 = 0(x - 0) \implies y = 4$$

   The line $y=4$ is a horizontal line. The distance from any point on this line to the X-axis (the line $y=0$) is 4 units. Since the origin $(0,0)$ lies on the X-axis, the distance from this tangent line to the origin is also 4 units. Proven.

5. Solution:

   Step 1: Determine the gradient from the angle.

   The relationship between the gradient ($m$) and the angle ($\theta$) a line makes with the positive X-axis is given by $m = \tan(\theta)$.

   $$m = \tan(45^\circ) = 1$$

   So, the gradient of the tangent line we are looking for is 1.

   Step 2: Find the coordinates of the point of tangency.

   The gradient is also the first derivative of the curve $f(x) = 2x^2 - 7x + 1$.

   $$f'(x) = 4x - 7$$

   We set it equal to the gradient we found:

   $$4x - 7 = 1$$

   $$4x = 8 \implies x = 2$$

   Now, find the $y$-value by plugging $x=2$ into the curve's equation:

   $$y = 2(2)^2 - 7(2) + 1 = 8 - 14 + 1 = -5$$

   So, the coordinates of the point of tangency are $(2, -5)$.

   Step 3: Determine the equation of the tangent line.

   Using the point $(2, -5)$ and gradient $m=1$:

   $$y - (-5) = 1(x - 2)$$

   $$y + 5 = x - 2$$

   $$y = x - 7$$